Electronic – Energy Transport in Circuits

circuit-theoryelectromagnetismenergyphysics

I've been reading about how energy is actually transported in circuits (I first read about it here: http://amasci.com/miscon/ener1.html). The description in this article of why electrons don't move around the circuit carrying energy like buckets, picking up energy from the battery and dumping it at the load, is reasonable to me, some points in particular being that electron drift velocity in circuits is actually very low and that in AC circuits electrons don't even drift along the entire length of the circuit but rather oscillate, so they don't even travel from sources to loads and back. Instead, it's electromagnetic waves that carry energy through a circuit.

I can accept this but now I'm confused about how you would interpret calculations of energy transfer in terms of electrons, calculations which also seem reasonable to me but seem to contradict the arguments above.

For example, suppose you have a 9 volt battery and you want to increase its energy by 500 J. Since a volt is J/C, \$ \Delta E = V \Delta C \rightarrow \Delta C = \frac{\Delta E}{V} \rightarrow \Delta C = \frac{500 J}{9 J/C} \approx 55.56\ Coul\$. Converting this into electrons, this is about \$ 3.47*10^{20}\$ electrons. This seems to suggest that you would physically need this number of electrons to pass between the terminals of the battery to transfer this energy. Is this calculation invalid? If not, how do you interpret this calculation in terms of electromagnetic waves?

Best Answer

This seems to suggest that you would physically need this number of electrons to pass between the terminals of the battery to transfer this energy. Is this calculation invalid? If not, how do you interpret this calculation in terms of electromagnetic waves?

With respect to your battery, when it provides one Coulomb of charge, oxidation in the anode really does cause 6E18 electrons to leave negative terminal, and a further 6E18 electrons to enter the positive terminal where they reduce a similar number of ions.

The point being made in your source is that it isn't the same electrons leaving one side as arriving at the other. Electrons are all interchangeable, and the ones going into the positive terminal probably started very, very close to where they ended up. You can think of electric field as pushing one electron a slight distance, which repulses another electron, which repulses another... until finally far down the chain one electron will reach the positive terminal and reduce an ion.

In addition, you can apply the same reasoning to the movement of ions in the battery cell itself, which also move under the effect of the cell voltage:

Battery cell diagram

Taken from wikipedia's article on galvanic cells.

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